A sample of 0.1     g of water at 100  o C and normal pressure 1.013 × 10 5  N  m - 2 requires 54     cal of heat energy to convert to steam at 100  o C . If the volume of the steam produced is 167.1     cc, the change in internal energy of the sample, is

Step-by-step solution

The values are given in the above question, ∆ Q = 54   c a l = 54 × 4.18   j o u l e = 225.72   j o u l e ∆ W = P V s t e a m - V w a t e r F o r   w a t e r   0.1   g r a m = 0.1  c c ∆ W = 1.013 × 10 5 167.1 × 10 - 6 - 0.1 × 10 - 6   j o u l e ⇒ ∆ W = 1.013 × 167 × 10 - 1 = 16.917   j o u l e . By first law of thermodynamics, the change in internal energy, ⇒ ∆ U = ∆ Q - ∆ W = 225 .72 -